Are the weak topologies of $\ell_1$ and $L_1$ homeomorphic?

Strangely may it sound, the question seeks contrasts between norm and weak topologies of Banach spaces from the non-linear point of view. And indeed, no linear homomorphism between weak topologies of non-isomorphic spaces exists as prevented by the Closed Graph Theorem. Note that the fact that $\ell_1$ is a Schur space yet $L_1$ is not is a priori not helpful, unless I am missing something trivial.

However, all infinite-dimensional, separable Banach spaces are homeomorphic. For weak topologies this is not so, as the weak topology of a reflexive space is $\sigma$-compact whereas this does not hold for $L_1$, say. We could then ask:

Is there a pair of non-isomorphic, non-reflexive Banach spaces whose weak topologies are homeomorphic?

31.03.2022: As noticed by Jerzy, the weak topologies of $\ell_1$ and $C[0,1]$ are not homeomorphic.

  • $\begingroup$ Is the Mazur map from $\ell_p$ to $\ell_2$ a weak homeomorphism when $1<p<\infty$? $\endgroup$ Mar 31 at 3:28
  • $\begingroup$ The space $l_1$ in the weak topology has a countable $k$-network (and having such a network is a topological property). The testing question might be if the space $L_1$ in the weak topology enjoys the same property. Note that the space $C[0,1]$ (which clearly contains a copy of $L_1$) in the weak topology does not admit a countable $k$-network (this is a result of Corson mentioned in Michal's paper). $\endgroup$ Mar 31 at 10:15
  • $\begingroup$ A network $\mathcal{N}$ in a topological space $X$ is called a $k$-network if for compact $K\subset X$ and open $U$ in $X$ with $K\subset U$ we have $K\subset\bigcup\mathcal{F}\subset U$ for some finite $\mathcal{F}\subset\mathcal{N}$, see E. Michael, $\aleph_{0}$-spaces, J. Math. Mech. 15 (1966) 983--1002. $\endgroup$ Mar 31 at 22:52

2 Answers 2


The weak topologies of the spaces $\ell_1$ and $L_1$ are not homeomorphic because of the following

Theorem. Assume that $X,Y$ are two Banach spaces whose weak topologies are homeomorphic. If $X$ has Shur's property, then $Y$ has Shur's property, too.

Proof. For a topological space $T$ let $T_s$ be the sequential coreflection of $T$, which is $T$ endowed with the topology consisting of sequentially open sets. A subset $U\subseteq T$ is sequentially open if for any convergent sequence $(x_n)_{n\in\omega}$ in $T$ with $\lim x_n\in U$ there exists $m\in\omega$ such that $x_n\in U$ for all $n\ge m$.

Assume that $h:X_w\to Y_w$ is a homeomorphism between the Banach spaces $X,Y$ endowed with the weak topologies. The homeomorphism $h$ remains a homeomorphism betweem the sequential coreflections $X_s$ and $Y_s$ of the spaces $X_w$ and $Y_w$, respectively. Since $X$ has Shur's property, the sequential coreflection $X_s$ of $X_w$ coincides with $X$. Since $X=X_s$ is a Baire space, so is the space $Y_s$. The Baireness of the space $Y_s$ implies that the closed unit ball $B_Y$ of $Y$ has non-empty interior in $Y_s$. Let $y_*$ be an interior point of the set $B_Y$ in $Y$. Assuming that $Y$ fails to have Shur's property, we can find a null sequence $(y_n)_{n\in \omega}$ in $Y_w$ consisting of vectors of norm $\|y_n\|=3$. Then $(y_n+y_*)_{n\in\omega}$ is a sequence in $Y_s\setminus B$ that converges to $y_*$, which is not possible as $y_*$ is an interior point of $B$ in $Y_s$. This contradiction implies that the Banach space $Y$ has Shur's property.

The proof of the above theorem suggests the following characterization of the Shur property.

Characterization. A Banach space $X$ has Shur's property if and only if the sequential coreflection $X_s$ of the weak topology of $X$ is a Baire space if and only if $X_s=X$.

Remark. Concerning the second question, it can be shown that any reflexive separable infinite-dimensional Banach spaces endowed with the weak topologies are sequentially homeomorhic (= the sequential coreflections of their weak topologies are homeomorphic), see this paper for more information in this direction.

  • $\begingroup$ +1 Professor Banakh, I'm excited to see your answer, for this is the first time I come across to the sequential coreflection that made a difficult problem look quite transparent. Could you suggest a reference about sequential coreflections of weak topologies? It is close to bounded-weak topology, but quite different if $X$ contains a copy of $\ell^1$. I wonder if DP-sets, limited sets, etc. could be characterized in this topology as well as the weak seq. continuity of multilinear maps & polynomials. My apologies for brevity and lack of foundation because of the limited space in the comments. $\endgroup$
    – Onur Oktay
    Apr 26 at 19:25
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    $\begingroup$ @OnurOktay Thank you for your comment. At the moment I do not have any good reference to the sequential coreflections of the weak topology (except for my paper mentioned in Remark at the end of my answer). Indeed, this topology coincides with the bounded-weak topology for Banach spaces with separable duals and also for reflexive Banach spaces. But for $\ell^1$ those two topologies are different. $\endgroup$ Apr 26 at 20:57
  • 1
    $\begingroup$ Thank you, Taras, very neat! $\endgroup$ Apr 26 at 21:22

Please read this as a long comment rather than a complete answer.

Let $X_1$ and $X_2$ be two (perhaps non-isomorphic) Banach spaces such that there exists a linear isomorphism $\phi:X_1^*\to X_2^*$. Let $\tau_i$ denote the weak topology of $X_i$. Let $\Phi:\tau_1\to\tau_2$ be the map that maps the subbases of the two weak topologies bijectively in the most natural way, and respects unions and finite intersections. That is, let $\Phi$ be defined as follows:

  • For $n\in\mathbb{N}$, for each $f_i\in X_1^*$ and $O_i\subseteq\mathbb{C}$ open $$\Phi\left(\bigcap_{i=1}^n f_i^{-1}(O_i)\right) = \bigcap_{i=1}^n(\phi f_i)^{-1}(O_i)$$
  • If $(G_{\alpha})_{\alpha\in I}$ is a family of open sets, each of which is of the form $\displaystyle \bigcap_{i=1}^n f_i^{-1}(O_i)$, then $$\Phi\left(\bigcup_{\alpha\in I}G_{\alpha}\right) = \bigcup_{\alpha\in I}\Phi\left(G_{\alpha}\right).$$

If $\Phi$ is a bijection, then $\Phi$ is an order isomorphism w.r.t. the inclusion of sets. Since both topologies are $T_1$, then $\Phi$ induces a homeomorphism between $\tau_1$ and $\tau_2$ (e.g. this and this).

  • $\begingroup$ What is the role of $\ell_1$ here? Also, example of what exactly is your construction? We know that the weak topologies of $\ell_1$ and $C[0,1]$ are not homeomorphic. $\endgroup$ Apr 19 at 13:56
  • $\begingroup$ Above $\ell^1$ is merely the common dual space for two linearly non-isomorphic Banach spaces, where there are plenty of those pairs math.stackexchange.com/questions/966208/… In case $\Phi$ is really bijective, then any two linearly non-isomorphic preduals of $\ell^1$ is a pair of non-isomorphic, non-reflexive Banach spaces whose weak topologies are homeomorphic. $\endgroup$
    – Onur Oktay
    Apr 19 at 19:41
  • $\begingroup$ Thanks, but since we know there exist such pair, the only question left is about $\ell_1$ and $L_1$. $\endgroup$ Apr 20 at 12:30
  • $\begingroup$ The dual spaces of $\ell^1$ and $L_1$ (that is $\ell^{\infty}$ and $L^{\infty}$) are isomorphic. Thus, if the proof above works for two preduals of $\ell^1$, it should also work for the pair $\ell^1$ and $L^1$ as well, so weak topologies of $\ell^1$ and $L^1$ must be homeomorphic. Nevertheless, it sounds to good to be true to me, perhaps there's a gap with the argument in the proof above. What is your insight? $\endgroup$
    – Onur Oktay
    Apr 20 at 22:44
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    $\begingroup$ Well, there exist Banach spaces $X$ and $Y$ with $X$ separable, $Y$ non separable, but $X^*$ and $Y^*$ are isomorphic; for example, $X=C[0,1]$ and $Y= C[0,1] \oplus c_0[0,1]$. Since the density of a Banach space is the same for a Banach space under the weak and norm topologies, $X$ and $Y$ are not weakly homeomorphic. $\endgroup$ Apr 21 at 0:37

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